[rho] x² + 1 tan² [delta] = ------- ([beta] - [alpha])-------------, (26) [sigma] 1/5 (1 - f^5) in which [sigma]/[rho] may be replaced by 800 times the S.G. of the metal, taking water as 800 times denser than air on the average, in round numbers, and formula (10) may be written n tan [delta] = [pi], or n[delta] = 180, when [delta] is a small angle, and given in degrees. From this formula (26) the table following has been calculated by A. G. Hadcock, and the results are in agreement with practical experience. 52. In the steady motion the centre of the shot describes a helix, with axial velocity / c1 \ u cos [theta]= v sin [theta] = ( l + -- tan² [theta] ) u cos [theta] [asympt] u sec [theta], (1) \ c2 / and transverse velocity / c1 \ u sin [theta] - v cos [theta] = ( l - -- ) u sin [theta] [asympt] ([beta] - [alpha]) u sin [theta]; (2) \ c2 / and the time of completing a turn of the spiral is 2[pi]/[mu]. When [mu] has the critical value in (7), 2[pi] 4[pi] C2 2[pi] ----- = ----- -- cos [theta] = ----- (x² + 1) cos [theta], (3) [mu] p C1 p which makes the circumference of the cylinder on which the helix is wrapped 2[pi] 2[pi]u -----(u sin [theta] - v cos [theta] = ------ ([beta] - [alpha]) (x² + 1) sin² [theta] cos [theta] [mu] p = nd ([beta] - [alpha]) (x² + 1) sin [theta] cos [theta], (4) and the length of one turn of the helix 2[pi] ----- (u cos [theta] + v sin [theta] ) = nd(x² + 1); (5) [mu] thus for x = 3, the length is 10 times the pitch of the rifling.